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Sedaia [141]
3 years ago
9

Which of the following is the greatest mass? Select one:

Chemistry
1 answer:
algol133 years ago
8 0

Answer:

B. 1.000 × 10-4 kg

Explanation:

A. 1.000 × 10-8 Mg = 0.00000001 Mg = 0.1 mg

B. 0.0001 kg = 0.1 g

C. 1000 ug = 1 mg

D. 1 .000 × 10-2 cg = 0.01 cg = 0.1 mg

A = D

B > C > A = D

You might be interested in
Select all possible answers:
Dafna11 [192]

Answer:

c and d are correct

Explanation:

In A, false because in Valence Electrons, the more the valences, the more stable an atom is.

In B, false because atoms cannot readily gain or lose valence electrons as the number of valence electrons is determined by the column they are in.

In C, true because the more the valence electrons, the more the stability of an atom.  

In D, true as electron placing is important and the reactivity of an atom is important.

So C and D are true!

7 0
3 years ago
Read 2 more answers
Take some points!!!​
VMariaS [17]

Answer: thanks babe

Explanation:

8 0
3 years ago
Read 2 more answers
An aqueous potassium carbonate solution is made by dissolving 5.51 moles of K 2 CO 3 in sufficient water so that the final volum
ad-work [718]

Answer:

1.67mol/L

Explanation:

Data obtained from the question include:

Mole of solute (K2CO3) = 5.51 moles

Volume of solution = 3.30 L

Molarity =?

Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:

Molarity = mole of solute /Volume of solution

Molarity = 5.51 mol/3.30 L

Molarity = 1.67mol/L

Therefore, the molarity of K2CO3 is 1.67mol/L

3 0
3 years ago
Read 2 more answers
Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W
nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

     H2= (1-x)Haq+XHvap.........1

    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

                        = 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

                     x = 1904.976 (J/g)/1998.576 (J/g)

                     x = 0.953

So, the quality of the wet steam is 0.953

                   

7 0
3 years ago
Read 2 more answers
How many moles of steam are produced when 5.74 moles of octane (C3H18) react?
jonny [76]

Answer:

51.66 mol H2O (steam)

Explanation:

5.74 mol C3H18 x 18 mol H2O/ 2 mol C3H18 = 51.66 mol H2O (steam)

6 0
3 years ago
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